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Question

A particle is executing SHM about y=0 along yaxis. Its position at an instant is given by y(m)=5(sin3πt+3cos 3πt). The amplitude of oscillation is
  1. 10 m
  2. 5 m
  3. 5(1+3) m
  4. 53 m

A
10 m
B
5(1+3) m
C
53 m
D
5 m
Solution
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y(m)=5(sin3πt+3cos3πt)y(m)=5×2(12×sin3πt+12×3cos3πt)y(m)=10(cosπ3×sin3πt+sinπ3×cos3πt)y(m)=10sin(3πt+π3)
So, amplitude of oscillation is 10.

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